# Is my function strongly convex?

Let's fix $$x \in \mathcal{X}$$ a finite discrete space . we consider $$F_{x}(h)=\epsilon \sum_{y}exp(-\frac{\Delta(x,y)}{\epsilon})$$ where $$\Delta(x,y) = \phi(x)+ \psi(y) + h(x)*(y-x) -c(x,y)$$ and $$\epsilon$$ is some fixed constant, phi and psi just 2 $$L^{1}$$ functions which do not matter here.

If we consider it's derivative with r.t h(x) we have $$F'_{x}(h)=-\sum_{y} exp(-\frac{\Delta(x,y)}{\epsilon})(y-x))$$ which i can't see why we can consider it strongly convex , i can consider it stricly convex at most but it doesnt give me the existence of a unique minimizer. I needed to apply Newton's algorithm to find a solution to $$F'_{x}(h)=0$$ for each $$x \in \mathcal{X}$$ but since the minimizer existence is not clear i cant really apply it.

• Without knowing $\phi$, $h$, and $c$ it is hard to even see that $F_x'$ is convex at all. By the way, strict convexity is sufficient for guaranteeing a unique minimizer (this is a classical exercise in convex analysis). Also, I'm not sure how you intend to apply Newton's method in a finite discrete place. Newton's method is an algorithm defined on a continuous space. Please elaborate.
– Zim
Jun 14, 2020 at 13:59
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